A290633 Lexicographically earliest sequence of positive integers such that, for any m and n > 0, gcd(a(n), a(n+1)) > 1 and a(n) != a(n+2), and if m < n then a(m) != a(n) or a(m+1) != a(n+1).

2, 2, 4, 4, 2, 6, 3, 3, 6, 2, 8, 4, 6, 6, 4, 8, 2, 10, 4, 12, 2, 14, 4, 10, 2, 12, 3, 9, 6, 8, 8, 6, 9, 3, 12, 4, 14, 2, 16, 4, 18, 2, 20, 4, 16, 2, 18, 3, 15, 5, 5, 10, 6, 12, 8, 10, 5, 15, 3, 18, 4, 20, 2, 22, 4, 24, 2, 26, 4, 22, 2, 24, 3, 21, 6, 10, 8, 12

Offset

1,1

Comments

a(n) > 1 for any n > 0.

If we drop the constraint "a(n) != a(n+2)", then we obtain the positive even numbers interspersed with 2's: 2, 2, 4, 2, 6, ...

Conjecturally, (a(n), a(n+1)) runs over all pairs of noncoprime positive integers; in this sense, this sequence is opposite to sequences like Stern's diatomic series (A002487).

This sequence has connections with A067992: here we avoid duplicate ordered pairs of consecutive terms, there unordered pairs, here we deal with noncoprime consecutive terms, there we (conjecturally) have coprime consecutive terms; also, the scatterplots of these sequences have similarities.

For any prime p, the sequence contains a multiple of p: by contradiction:

- let p be the least prime whose multiples are missing from the sequence (note that p > 2),

- there is only a finite number of pairs of noncoprime (p-1)-smooth numbers < p^2,

- so eventually we must have a term, say a(m), > p^2,

- if q is the least prime factor of a(m-1), then p*q would have been a better choice for a(m), hence the contradiction.

Also, if p is an odd prime, then the first multiple of p appearing in the sequence is a semiprime p*q with q < p.

If p < q are prime, then the first multiple of p appears before the first multiple of q.

For any prime p, the first occurrence of p in the sequence is immediately followed by a second occurrence of p.

For any prime p > 3:

- there is a semiprime p*q with q < p in the sequence,

- if q = 2, then this first p*q is followed by a 4,

- if q > 2, then this first p*q is followed by a 2,

- so there are infinitely many 2's or 4's in the sequence,

- if there are infinitely many 2's in the sequence, then the n-th occurrence of 2 is followed by 2*(n+e) with |e| <= 1, and every even

number appears in the sequence,

- the same conclusion applies if there are infinitely many 4's,

- hence every even number appear in the sequence.

For any n > 1, the first occurrence of n in the sequence must be either preceded or followed by the least prime factor of n (A020639).

Links

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Rémy Sigrist, PARI program for A290633

Rémy Sigrist, Scatterplot of the first 10000 pairs of consecutive terms

Rémy Sigrist, Colorized scatterplot of the first 100000 pairs of consecutive terms

Example

a(1) = 2 is suitable.
a(2) = 2 is suitable.
a(3) cannot be either 2 (=a(1)) or 3 (gcd(2,3)=1).
a(3) = 4 is suitable.
a(4) cannot be either 2 (=a(2)) or 3 (gcd(4,3)=1).
a(4) = 4 is suitable.
a(5) = 2 is suitable.
a(6) cannot be 2 (pair (2,2) already seen), 3 (gcd(2,3)=1), 4 (pair (2,4) already seen) or 5 (gcd(2,5)=1).
a(6) = 6 is suitable.

Prog

(PARI) See Links section.

Crossrefs

Cf. A002487, A020639, A067992.

Sequence in context: A056673 A353761 A128442 * A038674 A330639 A339179

Adjacent sequences: A290630 A290631 A290632 * A290634 A290635 A290636

Keyword

nonn,look

Author

Rémy Sigrist, Aug 08 2017

Status

approved